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Editorial Team
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Asked: 2026-06-16T21:29:05+00:00

Please have a look at the code below and explain to me why there

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Please have a look at the code below and explain to me why there is a deviance in the final results. Note that the difference is the introduction of the brackets in the second calculation. Thanks!

Code:

DECLARE  @A decimal(38,19) = 7958011.98
DECLARE  @B decimal(38,19) = 10409029441
DECLARE  @C decimal(38,19) = 10000000000

DECLARE  @Z1 decimal(38,19)
DECLARE  @Z2 decimal(38,19)


SET @Z1 = @A * @B / @C
SET @Z2 = @A * (@B / @C)

SELECT  @Z1 AS [Correct], 
        @Z2 AS [Wrong]

Results:

Correct = 8283518.0991650000000000000
Wrong   = 8283510.5860060000000000000
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1 Answer

  1. Editorial Team

    The intermediate datatypes are different because of this MSDN article

    That is, (@B / @C) evaluated first, follows rules like this. The intermediate datatype then affects the multiplication by @A

    You can see the intermediate and final types here (before assigning to a decimal(38,19) type

    SELECT
     @A * @B,        -- decimal (x, 6)
     @A * @B / @C,   -- decimal (x, 6)
     (@B / @C),      -- decimal (x, 6)
     @A * (@B / @C)  -- decimal (x, 6)

    So, instead of 1.0409029441 you get 1.040902 for your 2nd math

    Note, your 1st is wrong too. It is actually 8283518.099165070318

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